已知函数F(x)=sin6次方x+cos6次方X-1,求函数的最小正周期与单调递减区间

f(x)=(sinx)^6+(cosx)^6-1
=[(sinx)^2]^3+[(cosx)^2]^3-1
=[(sinx)^2+(cosx)^2][(sinx)^4-(sinxcosx)^2+(cosx)^4]-1
=(sinx)^4-(sinxcosx)^2+(cosx)^4-1
=[(sinx)^2+(cosx)^2]^2-3(sinxcosx)^2-1
=1-3(sinxcosx)^2-1
= -3(sinxcosx)^2
= -3[1/2*2sinxcosx]^2
= -3*1/4*[sin(2x)]^2
= -3/4*[1-cos(4x)]/2
=3/8*cos(4x)-3/8 ,
因此函数最小正周期为 2π/4=π/2 ,
单调递减区间是 [kπ/2,kπ/2+π/4] ,k∈Z .